[tex]\frac{-b+-\sqrt{b^2-4ac} }{2a}; ax^2+bx+c \\ \\ a=1, b=-12, c=59[/tex]
Use the formulas above to solve the equation.
[tex]1x^2-12x+59=0\\ \\ \frac{-(-12)+-\sqrt{(-12)^2-4(1)(59)} }{2(1)}\\ \\ \frac{12+-\sqrt{144-236} }{2}\\ \\ \frac{12+-\sqrt{-92} }{2}[/tex]
Now this problem will be giving i values since there's a negative in the square root. These i's are known as imaginary numbers. For the answer it should be these two:
x =(12-[tex]\sqrt{-92}[/tex])/2=6-i[tex]\sqrt{23}[/tex] = 6.0000-4.7958i
x =(12+[tex]\sqrt{-92}[/tex])/2=6+i[tex]\sqrt{23}[/tex] = 6.0000+4.7958i
Let me know if you need help with finding i and all or if you just needed the answer to check your work. Hope this helps!
